Organic Chemistry (8th Edition)

Published by Pearson
ISBN 10: 0321768418
ISBN 13: 978-0-32176-841-4

Chapter 4 - The Study of Chemical Reactions - Study Problems - Page 171: 4-40 b

Answer

In kJ/mol: -17kj/mol In kcal/mol: -4kcal/mol

Work Step by Step

In the given reaction, bond breaking and bond formation takes place. Bond breaking is endothermic (positive) whereas bond formation is exothermic. In order to find $\Delta$ $H^{\circ}$ of the reaction, we add the bond dissociation energies, keeping in mind that where the bond formation takes place, the bond dissociation energy is negative. Therefore, In kJ/mol Bonds broken= $CH3CH2-Cl=+339$ $H-I= +297$ Bonds formed= $CH3CH2-I=-222$ $H-Cl=-431$ $\Delta$$H^{\circ}$= $(+339+297) + (-222) + (-431)$ $\Delta$$H^{\circ}$= $+636+(-653)$ $\Delta$$H^{\circ}$= $+636 -653$ $-17 kJ/mol$ In kcal/mol Bonds broken= $CH3CH2-Cl= +81$ $H-I= +71$ Bonds formed= $CH3CH2-I=-53$ $H-Cl=-103$ $\Delta$$H^{\circ}$=$ (+81+71 )+ (-53)+(-103)$ $\Delta$$H^{\circ}$ =$152+ ( -156)$ $\Delta$$H^{\circ}$ = $152 - 156$ $\Delta$$H^{\circ}$= $-4 kcal/mol$
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